Answer :
Let B =
To apply elementary row transformations we write:
B = IB where I is the identity matrix
We proceed with operations in such a way that LHS becomes I and the transformations in I give us a new matrix such that
I = XB
And this X is called inverse of B = B-1
So we have:
Applying R2→ R2 + 2R1
⇒
As we got all zeroes in one of the row of matrix in LHS.
So by any means we can make identity matrix in LHS.
∴ inverse of B does not exist.
B-1 does not exist. …ans
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